According to the statement, we have the following balanced reaction:
[tex]BaCl_{2(aq)}+Na_2SO_{4(aq)}\rightarrow BaSO_{4(s)}+2NaCl_{(aq)}[/tex]We see that the ratio BaCl2 to Na2SO4 is 1/1, that is, the same amount of moles of both reactants are needed to react completely. Now, to find the normality of the solution we will follow the following steps.
1. We find the moles of Na2SO4 by dividing the given mass by its molar mass.
2. We find the moles of BaCl2 by the stoichiometry of the reaction.
3. We apply the normality equation. Normality indicates the amount of solute equivalents in 1 liter of solution, it is very similar to molarity, but here we will do a little conversion.
Let's proceed with the calculation.
1. Moles of Na2SO4
[tex]\begin{gathered} molNa_2SO_4=givengNa_2SO_4\times\frac{1molNa_2SO_4}{MolarMass,gNa_2SO_4} \\ molNa_2SO_4=554mgNa_2SO_4\times\frac{1g}{1000mg}\times\frac{1molNa_2SO_4}{142.04gNa_2SO_4}=3.90\times10^{-3}molNa_2SO_4 \end{gathered}[/tex]2. Moles of BaCl2
It will be the same moles of Na2SO4, so moles of BaCl2 will be 3.90x10^-4 mol
3. Normality of the solution
It is known that in 1 mol of BaCl2 there are 2 equivalents. So, we have:
[tex]N=\frac{3.9\times10^{-4}molBaCl_2}{54.8mL}\times\frac{1000mL}{1L}\times\frac{2eq}{1molBaCl_2}=0.142N[/tex]The normality of the solution will be: 0.142N
